若π/4<α<3π/4,0<β<π/4,cos(π/4-α)=3/5,sin(3π/4+β)=5/13...

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∵π/4＜α＜3π/4 ∴-π/2<π/4-α<0 ∵cos(π/4-α)=3/5 ∴sin(π/4-α)=-4/5 ∵0＜β＜π/4 ∴3π/4<3π/4+β<π ∵sin（3π/4+β）=5/13 ∴cos(3π/4+β)=-12/13 ∴sin(α+β)=-cos(π/2+α+β)=-cos[(3π/4+β)-(π/4-α)]=-cos(3π/4...

cos(π/4-α)=3/5，则sin(π/4-α)<0 则sin(π/4-α)=-根号[1-（cos(π/4-α)）^2]=-4/5 3π/4<3π/4+β<π sin(3π/4+β)=5/13 则cos(3π/4+β)<0 cos(3π/4+β)=-12/13 而sin(α+β)=-cos（3π/4+β-（π/4-α））=-cos（α+β+π/2）而 cos...

sin（α+β）= cos(π/4+α+π/4+β)= [cos（π/4+α）*cos（π/4+β）－sin（π/4+α）*sin（π/4+β）]=－[(√(1-9/25)*(5/13)－(3/5)√(1-25/169)]=97/352

cos(π/4-a)=3/5 π/4<a<3π/4 π/4-a<0 sin(π/4-a)=-4/5 sin(3π/4+b)=5/13 0<b<π/4 3π/4<3π/4+b<π cos(3π/4+b)=-12/13 sin(a+b)= sin[（3π/4+b）-（π/4-a)-π/2]=-cos[（3π/4+b）-（π/4-a)]=-[cos（3π/4+b）cos(π/4-...

π/4＜a＜3π/4 , π/2 <π/4+a<π 当2kπ<x<π+2kπ 时 sinx>0 ,0＜β＜π/4 3π/4 <3π/4+β<π 当 π/2 + 2kπ<x<3π/2+2kπ 时 cosx<0

sin(¾π+α)=sin[π-(¾π+α)]=sin(¼π-α)=5/13 0<α<π/4→α=¼π-arcsin(5/13)cos(¼π-β)=cos(β-¼;π)=3/5→β=¼π+arccos(3/5)cos(α+β)=cos[½π-(arcsin(5/13)-arccos(3/5))]=sin[arcsin(5/13)-arc...

∴π/4<π/4+α<π/2，-π/2<π/4-β<0 ∴sin(π/4+α)=12/13，sin（π/4-β）=-4/5 则cos（α+β）=cos[(π/4+α)-(π/4-β)]=cos(π/4+α)cos(π/4-β)+sin(π/4+α)sin(π/4-β)=(5/13)*(3/5)+(12/13)*(-4/5)=15/65-48/65 =-33/65 ...

又π/4＜a＜3π/4， -π/2＜π/4-a＜0 ∴sin（π/4-a）=-4/5 同理：sin(3π/4+b)=5/13；cos（3π/4+b）=-12/13 注意：a+b=[（3π/4+b）-（π/4-a）]-π/2 ∴sin（a+b）=sin{[（3π/4+b）-（π/4-a）]-π/2}=-cos{[（3π/4+b）-（π/4-a）]=...

0.π/2））sin(3π/4+β)=5/13，cos（3π/4+β)=-12/13（因为β∈(0,π/4，所以3π/4+β∈(3π/4，π））cos(α-π/4)sin(3π/4+β)+cos（3π/4+β)sin(α-π/4)=sin（α-π/4+3π/4+β）=sin（α+β+π/2）=-33/65 sin（α+β)=56/65 望采纳 ...